Real Analysis

The Well-Ordering Property

The well-ordering property of \(\mathbb{N}\): every non-empty subset of \(\mathbb{N}\) has a least element. It can be taken as an axiom of \(\mathbb{N}\).

Principle of Induction

Let \(S \subset \mathbb{N}\) such that

\(1 \in S\), and

if \(k \in S\) then \(k + 1 \in S\).

Then \(S = \mathbb{N}\).

The well-ordering property is equivalent to the principle of induction.

Proof that the well-ordering property implies the principle of induction (by contradiction). Suppose \(S\) exists with the given properties but \(S \neq \mathbb{N}\).

Let \(A := \mathbb{N} \setminus S\), so it is non-empty. Then \(A\) has a least element by the well-ordering property, call it \(n\). Notice \(n > 1\) because \(1 \in S\) which implies \(1 \notin A\).

Consider \(n - 1\). It is not in \(A\) because \(n\) is the least element. So \(n - 1 \in S\). By property (2), \(n - 1 + 1 \in S\), which implies \(n \in S\). This is a contradiction because \(n \in A\).

Therefore, the well-ordering property implies the principle of induction. QED.

Proof that the principle of induction implies the well-ordering property (by induction on the number of elements in the subset). Let \(S\) be a subset of \(\mathbb{N}\) and \(n\) be the number of elements in \(S\).

For the base case where \(n = 1\), the least element of any subset \(S\) is the only element it has, so the base case holds.

For the inductive step, assume for all subsets \(S\) of \(\mathbb{N}\) with size \(n\), each has a least element. Consider a subset \(S\) of \(\mathbb{N}\) with size \(n + 1\) and a subset \(S'\) of \(S\) with size \(n\). By the inductive hypothesis, \(S'\) has a least element; call it \(a\). Let \(b\) is the only element in \(S\) that is not in \(S'\). Since \(\mathbb{N}\) is ordered, either \(a < b\), \(a = b\), or \(a > b\), by law of trichotomy. When \(a < b\), \(a\) is a least element of \(S\). When \(a = b\), \(a\) is also a least element of \(S\). When \(a > b\), \(b\) is less than all elements of \(S'\), so \(b\) is a least element of \(S\).

Therefore, the principle of induction implies the well-ordering property. QED.

Proof by Induction

Let \(P(n)\) be statements indexed by the natural numbers \(\mathbb{N}\). To show that \(P(n)\) is true for all \(n\), we must show

(base case) \(P(1)\) is true, and

(inductive step) if \(P(k)\) is true then \(P(k + 1)\) is true.

Then, by the principle of induction, \(P(n)\) is true for all \(n\).

The statement \(P(k)\) is true is called the inductive hypothesis.

Strong Induction

Let \(P(n)\) be statements indexed by the natural numbers \(\mathbb{N}\). To show that \(P(n)\) is true for all \(n\), we must show

(base case) \(P(1)\) is true, and

(inductive step) if \(P(1)\) through \(P(k)\) are true then \(P(k + 1)\) is true.

Then, by the principle of strong induction, \(P(n)\) is true for all \(n\).

For example, every \(2^n \times 2^n\) chessboard with one square removed can be tiled by L-shaped tiles.